Question

If x^2+y^2= 27xy then show that log (x-y/5) = 1/2 (log^x + logy )
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Answer 1

Answer:

Step-by-step explanation:

given

x^2+y^2= 27xy

x^2+y^2-2xy= 25xy

(x-y)^2=25xy

(x-y)=(25xy)^1/2

(x-y)=5(xy)^1/2

(x-y)/5=(xy)^1/2

Now take log on both sides, we get

log(x-y/5) = log(xy) ^1/2

log​(x-y/5) = 1/2 log(xy)

​log(x-y/5) = 1/2 (logx + logy)

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